Codeforces Round #427 (Div. 2) D. Palindromic characteristics(Manacher求回文串)

 1 #include<bits/stdc++.h>
 2 #define F(i,a,b) for(int i=a;i<=b;++i)
 3 using namespace std;
 4 
 5 const int N=5000+7;
 6 
 7 char s[N];
 8 int ans[N],n;
 9 struct Manacher{
10     char str[N<<1];
11     int p[N<<1],len,mx,id,tl,ans,i;
12     int maxlen(char *s){
13         len=strlen(s),mx=0,id=0,tl=0,str[tl++]=‘$‘,str[tl++]=‘#‘;
14         for(i=0;i<len;i++)str[tl++]=s[i],str[tl++]=‘#‘;
15         for(i=2,str[tl]=0,ans=0;i<tl;i++){
16             p[i]=mx>i?min(p[(id<<1)-i],mx-i):1;
17             while(str[i-p[i]]==str[i+p[i]])p[i]++;
18             if(i+p[i]>mx)mx=i+p[i],id=i;
19             if(ans<p[i])ans=p[i];
20         }
21         return ans-1;
22     }
23 }M;
24 
25 int is[N][N];
26 
27 int main()
28 {
29     scanf("%s",s);
30     M.maxlen(s);
31     F(i,1,M.len)is[i][i]=1;
32     ans[1]=M.len;
33     F(i,1,M.len*2)
34     {
35         if(M.p[i]<2)continue;
36         int R;
37         R=M.p[i]/2;
38         if(M.str[i]!=‘#‘)
39         {
40             int idx=i/2;
41             for(int j=R-1;j>=1;j--)
42             {
43                 int l=idx-j,r=idx-1; 
44                 is[l][idx+j]=is[l][r]+1;
45                 ans[is[l][idx+j]]++;
46             }
47         }else
48         {
49             int idx=i/2;
50             for(int j=R;j>=1;j--)
51             {
52                 int l=idx-j+1,r=idx;
53                 is[l][idx+j]=is[l][r]+1;
54                 ans[is[l][idx+j]]++;
55             }
56         }
57     }
58     for(int i=M.len-1;i;i--)ans[i]+=ans[i+1];
59     F(i,1,M.len)printf("%d%c",ans[i]," \n"[i==M.len]);
60     return 0;
61 }